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Mandelbrot set 🌀

Info

This is just for fun, there's no real-world application for this (as far as I know) 😝

Question

Plot an image of the Mandelbrot set.

Start with a 51x51 grid of points ranging from -2 to 2 on both the x and y axes. A sample of some points in the grid are:
(-2, 2) (-1.92, 2) ... (1.92, 2) (2, 2)
(-2, 1.92) (-1.92, 1.92) ... (1.92, 1.92) (2, 1.92)
... ... (0, 0) ... ...
(-2, -1.92) (-1.92, -1.92) ... (1.92, -1.92) (2, -1.92)
(-2, -2) (-1.92, -2) ... (1.92, -2) (2, -2)

Apply the Mandelbrot set formula to each point in the grid for 100 iterations. If the point remains inside this grid after 100 iterations, mark it with a •. If it diverges, mark it with a space.

Once you know which points are in the set, return a single cell with a string representation of the grid.

For example, an output might look like the following string:

         •
        •
        ••
      •••••
    • •••••
••••••••••••
    • •••••
      •••••
        ••
        •
         •

The Mandelbrot set is a set of complex numbers c for which the function f(z) = z^2 + c does not diverge when iterated from z = 0.

We can plot the Mandelbrot set by considering the complex plane as a grid of coordinates. Given two points (a, b) and (c, d), we can define addition and multiplication as:

  • Addition: (a, b) + (c, d) = (a + c, b + d)
  • Multiplication: (a, b) * (c, d) = (a * c - b * d, a * d + b * c)

Note that, by considering the complex plane as a grid of coordinates, the complex number 0 is represented as (0, 0).

The solution can be found at:

A worked example is provided below to help illustrate the Mandelbrot set calculations.

Sample input

Plot an image of the Mandelbrot set on a 21x21 grid.

Sample output 
          •
         •
         ••
       •••••
     • •••••
••••••••••••
     • •••••
       •••••
         ••
         •
          •

solution(mandelbrot_set) as (
    select concat_ws(e'\n',
        '                     ',
        '                     ',
        '                     ',
        '                     ',
        '                     ',
        '          •          ',
        '         •           ',
        '         ••          ',
        '       •••••         ',
        '     • •••••         ',
        '••••••••••••         ',
        '     • •••••         ',
        '       •••••         ',
        '         ••          ',
        '         •           ',
        '          •          ',
        '                     ',
        '                     ',
        '                     ',
        '                     ',
        '                     '
    )
)
Hint 1

Generate the grid of points into a table with columns x and y ranging from -2 to 2 in steps of 0.08.

Hint 2

Use a recursive CTE to iterate over the grid points for 100 iterations, applying the Mandelbrot set formula to each point.

Worked examples

To help illustrate the Mandelbrot set calculations, consider the following points:

  • (-2, 2)
  • (-2, 0)
  • (-1, 0)
  • (0, 0)
  • (2, 0)
  • (1.04, 1.04)

Let's walk through the steps for these.

We'll use two additional details:

  1. For a complex number z = (a, b), the square of z is z^2 = (a^2 - b^2, 2ab)
  2. Any complex number that exceeds -2 or 2 on either axis will diverge

(-2, 2)

In the function f(z) = z^2 + c, we start with z = (0, 0) and c = (-2, 2). The first two iterations are:

  • f((0, 0)) = (0, 0)^2 + (-2, 2) = (-2, 2)
  • f((-2, 2)) = (-2, 2)^2 + (-2, 2) = (4 - 4, 2 * -2 * 2) + (-2, 2) = (0, -8) + (-2, 2) = (-2, -6)

After the second iteration, the resulting point is outside the grid, so it will diverge; therefore, the point (-2, 2) is not in the set.

(-2, 0)

In the function f(z) = z^2 + c, we start with z = (0, 0) and c = (-2, 0). The first few iterations are:

  • f((0, 0)) = (0, 0)^2 + (-2, 0) = (-2, 0)
  • f((-2, 0)) = (-2, 0)^2 + (-2, 0) = (4 - 0, 2 * -2 * 0) + (-2, 0) = (4, 0) + (-2, 0) = (2, 0)
  • f((2, 0)) = (2, 0)^2 + (-2, 0) = (4 - 0, 2 * 2 * 0) + (-2, 0) = (4, 0) + (-2, 0) = (2, 0)

After the second iteration, the resulting point remains the same and within the grid; therefore, the point (-2, 0) is in the set.

(0, 0)

In the function f(z) = z^2 + c, we start with z = (0, 0) and c = (0, 0). The first iteration is:

  • f((0, 0)) = (0, 0)^2 + (0, 0) = (0, 0)

The resulting point remains the same after the first iteration, so it will not diverge; therefore, the point (0, 0) is in the set.

(2, 0)

In the function f(z) = z^2 + c, we start with z = (0, 0) and c = (2, 0). The first two iterations are:

  • f((0, 0)) = (0, 0)^2 + (2, 0) = (2, 0)
  • f((2, 0)) = (2, 0)^2 + (2, 0) = (4 - 0, 2 * 2 * 0) + (2, 0) = (4, 0) + (2, 0) = (6, 0)

After the second iteration, the resulting point is outside the grid, so it will diverge; therefore, the point (2, 0) is not in the set.

(1.04, 1.04)

In the function f(z) = z^2 + c, we start with z = (0, 0) and c = (1.04, 1.04). The first two iterations are:

  • f((0, 0)) = (0, 0)^2 + (1.04, 1.04) = (1.04, 1.04)
  • f((1.04, 1.04))
    = (1.04, 1.04)^2 + (1.04, 1.04)
    = (1.0816 - 1.0816, 2 * 1.04 * 1.04) + (1.04, 1.04)
    = (0, 2.1632) + (1.04, 1.04)
    = (1.04, 3.2032)

After the second iteration, the resulting point is outside the grid, so it will diverge; therefore, the point (1.04, 1.04) is not in the set.